Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{-q - 3}{q^2 - 5q - 14} \times \dfrac{-8q - 16}{2q + 6} $
Explanation: First factor the quadratic. $a = \dfrac{-q - 3}{(q + 2)(q - 7)} \times \dfrac{-8q - 16}{2q + 6} $ Then factor out any other terms. $a = \dfrac{-(q + 3)}{(q + 2)(q - 7)} \times \dfrac{-8(q + 2)}{2(q + 3)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ -(q + 3) \times -8(q + 2) } { (q + 2)(q - 7) \times 2(q + 3) } $ $a = \dfrac{ 8(q + 3)(q + 2)}{ 2(q + 2)(q - 7)(q + 3)} $ Notice that $(q + 3)$ and $(q + 2)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 8(q + 3)\cancel{(q + 2)}}{ 2\cancel{(q + 2)}(q - 7)(q + 3)} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $a = \dfrac{ 8\cancel{(q + 3)}\cancel{(q + 2)}}{ 2\cancel{(q + 2)}(q - 7)\cancel{(q + 3)}} $ We are dividing by $q + 3$ , so $q + 3 \neq 0$ Therefore, $q \neq -3$ $a = \dfrac{8}{2(q - 7)} $ $a = \dfrac{4}{q - 7} ; \space q \neq -2 ; \space q \neq -3 $